My exercise is asking to prove the following:
$ \int_{0}^{t} W_s^2 \ dW_S = \frac{1}{3}W_t^3 - \int_{0}^{t} W_sds $
I have applied Ito's formula here, using a defined process $f = X^3$, computed the time derivative and the first and second space derivative and arrived at the following solution:
$dX^3 = 3xdt + 3x^2dX_t$
I think there is something trivial I am missing here, but I am a bit unsure about how the integration would work to lead me to the result I am expected to observe and any pointers would be appreciated.
The notation $$ dX_t^3 = 3 X_t dt + 3 X_t^2 dX_t $$ is (by its very definition) only another way of writing $$ X_t^3 - X_0^3 = \int_0^t 3 X_t dt + \int_0^t 3 X_t^2 dX_t. $$ So you are actually done.